JEE PHYSICS MCQ: SYSTEM OF PARTICLES

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Application +4 | -1

An astronaut is rotated in a horizontal circle at a radius of

5 \mathrm{~m}

. If the astronaut can withstand a maximum acceleration of

7 \mathrm{~g}

, then maximum numbers of permissible revolution per second is

\left(1 \mathrm{~g}=10 \mathrm{~m} / \mathrm{s}^2\right)

View answer and explanation

Correct option: A

\frac{\sqrt{14}}{2 \pi} \frac{\mathrm{rev}}{\mathrm{s}}

Option A (

\frac{\sqrt{{14}}}{{2 \pi}} \frac{{\mathrm{{rev}}}}{{\mathrm{{s}}}}

) is correct because it represents the maximum permissible revolutions per second calculated from the maximum allowable acceleration.

More Questions from SYSTEM OF PARTICLES

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\mathrm{AB}

\mathrm{L}

\mu(x)=a+\frac{b x}{L}

x

\mathrm{A}

\mathrm{CM}

\left(\frac{7}{12}\right) \mathrm{L}

\mathrm{A}

\mathrm{b}

1.0 \mathrm{~g}

8 \mathrm{~cm}

\omega = {t^4} + 2{t^3}

t = 1\,sec

\left( {rad\,{s^{ - 2}}} \right)